翻转链表二

Reverse a linked list from position m to n.

 Notice

Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.

Have you met this question in a real interview? 

Yes

Example

Given 1->2->3->4->5->NULL, m = 2 and n = 4, return1->4->3->2->5->NULL.

要注意是否从第一个节点开始翻转，1）是则保存第n+1个节点，进行翻转，然后返回s2）否则保存第m-1个节点，和第n+1个节点，翻转

/**
 * Definition of singly-linked-list:
 * 
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *        this->val = val;
 *        this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The head of linked list.
     * @param m: The start position need to reverse.
     * @param n: The end position need to reverse.
     * @return: The new head of partial reversed linked list.
     */
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        // write your code here
        int i = m;
        int j = n - m;
        ListNode *p = head;
        int count = 0;
        while(p != NULL)
        {
            p = p->next;
            count++;
        }
        if(count < n)return NULL;
        p = head;
        while(i > 2)
        {
            p = p->next;
            i--;
        }
        ListNode *bf = p;
        if(m != 1)p = p->next;
        ListNode *q = p;
        while(j >= 0)
        {
            q = q->next;
            j--;
        }
        ListNode *s = q;
        while(p != q)
        {
            ListNode *tmp = p->next;
            p->next = s;
            s = p;
            p = tmp;
        }
        if(m != 1)
        {
            bf->next = s;
            return head;
        }
        else return s;
        
    }
};